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Shaft Design

ENTC 463Mechanical Design Applications II

Next Thursday4/3/2008

Meet @ Thompson 009B

ENTC 463Mechanical Design Applications II

Allen to Hagan – 2:20 to 3:00 PMHigginbotham to Winniford– 3:00 to 3:35 PM

ATTENTION: MMET Students, the IAC members would like to meet YOU on April 11th !

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Deadline to sign up is April 8th.

MMET IAC MeetingApril 11th 2008

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Sign up with Courtney in THOM 117

ENTC 463

• Lab 3 Due 4/10• Homework HW 12 Due 4/15

– Chapter 12 – 2, 4, 25, 27

Shafts

• Mott, Chapter 12• Why use shaft?

– To transmit power• Shaft geometry

– Cylinder, bar, beam (length and diameter)• Load acting on shaft

– Torsion (shear stress)– Bending (normal stress)

Shaft Design

Shaft Design

Shaft Design

• Given required power to be transmitted– Calculate torque, – Calculate forces, – Calculate stresses (if

geometry is known),– Select material

• Given required power to be transmitted– Calculate torque, – Calculate forces, – Determine shaft

diameter (if the material is known)

Shaft Design Procedure1. Develop the free body diagram; model the

various machine elements mounted on the shaft in terms of forces and torques

2. Develop the shear and moment diagram; identify bending moment (leads to normal stress) and torque (leads to shear stress)

3. Identify critical locations for stress analysis; calculate stresses (known diameter)

4. Determine diameter or select material based on failure theories

Forces Acting on Shaft

• Forces due to gear (spur gear)

gear) (helical tantan2

63000

ψφ

tx

tr

t

WWWWDTW

nhpT

==

=

=

Forces due to Gears

Forces Acting on Shaft

• Forces due to chain and sprocket

θθ

sincos

222

ccy

ccx

B

B

A

Ac

FFFF

DT

DT

DTF

==

===

Forces Acting on Shaft

• Forces due to V-belt and sheave

20.2

pulley andbelt flat For 2

5.121

DTF

DTF

FFF

B

B

B

−=

Example

• A chain is transmitting 100 kW with the chain speed at 6000 rpm and V = 50 m/s. The shaft material is AISI 1040 cold drawn. Determine the shaft diameter required.

Shaft Design Considerations

• Stress Concentration (fillet or key seat)1.5 < Kt < 2.5

• Combined tangential and radial load (3-D)– Two shear and moment diagrams

Wt

Wr

22yzxyy MMM +=

x

z

y

Stress Concentration

• Keyseats– Kt = 2.0 for profile keyseat– Kt = 1.6 for sled keyseat

• Shoulder fillets– Kt = 2.5 for sharp fillet– Kt = 1.5 for well-rounded fillet

• Retaining ring grooves– Kt = Kt = 3.0, or– Increase diameter by 6%

Forces Acting on A Shaft

Shear and Moment DiagramsFrom bottom look up Front view

Shaft Design/Analysis Example

( ) ( )( )

31

22

31

22

223

223

223

31

32

32

32

1616

:

⎟⎟⎟

⎜⎜⎜

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛≥

⎟⎟⎠

⎞⎜⎜⎝

⎛+≥

≤+

≤+−−++

≤−

yy

y

y

y

y

ST

SMNd

TMSNd

NS

TMd

NS

TMMd

TMMd

NS

MSST

π

π

π

ππ

σσ

( )( )( )

( )

( )

( )223

22

1

223

22

1

34

34

1622

1622

16322

32642

TMMd

TMMd

dT

ddT

JTc

dM

ddM

IMc

xyxx

xyxx

xy

x

+−=+⎟⎠⎞

⎜⎝⎛−=

++=+⎟⎠⎞

⎜⎝⎛+=

===

===

πτσσσ

πτσσσ

ππτ

ππσ

Is this correct ?

Fatigue Failure Criterion

• Cyclic loading due to shaft rotation– Find mean and alternating stresses– Construct Mohr’s circles for mean stress and

alternating stress– Derive effective mean and alternating

stresses (based on MSST or DET)– Use Soderberg or Goodman for design and

analysis

Fatigue Failure of Shaft

3

3

16

0

32

dT

dM

xy

y

x

πτ

σπ

σ

=

=

±=

3

16

00:Mean

dT

mxy

my

mx

πτ

σσ

=

==

0

0

32:gAlternatin

3

=

=

=

mxy

my

ax dM

τ

σπ

σ

NSSK

dMdT

y

m

n

at

a

m

1''

':Soderberg

32'

32'

3

3

=+

=

=

σσ

πσ

πσ

1'

:sionsteady tor and bending reversedfully for equation ANSI/ASME

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

ys

m

n

a

ANSI Shaft Design Equation

For repeated reversed bending and constant torque

ANSI/ASME Shaft Equation

1'

:sionsteady tor and bending reversedFully 22

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

ys

m

n

a

1'

1'

1'

:factorsafety Consider

2222

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛⇒=⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛⇒

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

ys

m

n

a

ys

m

n

a

ys

m

n

a

SNτ

SNσ

NSτ

NSσ

ANSI / ASME Shaft Equation

13'

31'

1'

:n)for torsioshear (pure DET Use

22

2

222

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛⇒

⎟⎟⎟⎟

⎜⎜⎜⎜

+⎟⎟⎠

⎞⎜⎜⎝

⎛⇒=⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

y

m

n

a

y

m

n

a

ys

m

n

a

SNτ

SNσ

S

NτSNσ

SNτ

SNσ

13'

13'

:ionconcentrat stressConsider 2222

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛⇒=⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

y

m

n

at

y

m

n

a

SNτ

SNσK

SNτ

SNσ

ANSI / ASME Shaft Equation

1

163

'

32

13'

16

32

2

3

2

322

3

3

=

⎟⎟⎟⎟

⎜⎜⎜⎜

⎛⎟⎠⎞

⎜⎝⎛

+

⎟⎟⎟⎟

⎜⎜⎜⎜

⎛⎟⎠⎞

⎜⎝⎛

⇒=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛⇒

==

==

yn

t

y

m

n

at

m

a

S

NdT

S

NdMK

SNτ

SNσK

dT

JTc

dM

IMcσ

ππ

πτ

π

31

22

43

'32

⎥⎥⎥

⎢⎢⎢

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛=

yn

t

ST

SMKNd

π

ANSI / ASME Shaft Equation

31

22

43

'32

sionsteady torandbendingreversedFully

⎥⎥⎥

⎢⎢⎢

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛≥

yn

t

ST

SMKNd

π

31

22

4332

Static

⎥⎥⎥

⎢⎢⎢

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛≥

yy ST

SMNd

π

ANSI Shaft Design Equation

31

22

43

'32

sionsteady torandbendingreversedFully

⎥⎥⎥

⎢⎢⎢

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛≥

yn

t

ST

SMKNd

π

1'

:sionsteady tor and bending reversedFully 22

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

ys

m

n

a

Can we use Soderberg or Goodman criterion?

Example 12-1 (p. 548)

• The system transmitting 200 hp from pinion P to gear A, and from pinion C to gear Q.

• The shaft rotating speed is 600 rpm.• Shaft material is AISI 1144 OQT 1000

Example (p. 549)• Free body diagram• Shear and moment diagrams• Torque at each segment• Calculate diameter for locations A, B, C, and D

(at both left and right)

A B C D

No momentTorque = 21000

No torque, no moment, vertical shear onlyNo torque

Shear and Moment DiagramsFrom bottom look up Front view

Design Examples

Design Example - Torque

Design Example - Forces

Design Examples – Shear and Moment